•  
• EDN Europe MAGAZINE
•  
• Free SUBSCRIPTION
• Subscription Renewal
• View Media Kit

• EDN
• EDN Australia
• EDN Asia
• EDN China
• EDN Japan

Resistor network sets gain for fixed-gain differential amplifier

When a fixed-gain amplifier doesn’t offer the optimum gain for your application, you can adjust the gain to a lower value by adding an external resistor network. This attenuation circuit works like a voltage divider but with a key difference: Resistors inside the fixed-gain amplifier load down the external network (Figure 1). For the differential-input configuration, you can reduce this system to an equivalent half-circuit for analysis (Figure 2).

You determine the voltage gain for the differential configuration as follows:

where gain is the amplifier’s fixed voltage gain in decibels. Maxim’s (www.maxim-ic.com) MAX9705 differential- audio amplifier, for example, is available in fixed-gain versions of 6, 12, 15.5, and 20 dB, and its input resistance is typically 20 kΩ. For singleended configurations, the gain is:

These gain equations assume that the frequency is much higher than the cutoff frequency of the highpass filter comprising C_IN and the equivalent input resistance of the circuit.

The fixed gain of the MAX9705 amplifier is typically within 5%, but the internal input resistors have an absolute tolerance of ±40%, which you must consider when calculating the resulting system gain. You must also account for the tolerance of the external resistors, using a worst-case approach for calculating the gain tolerance. Replacing the resistor values with their maximum deviations from normal—that is, the extreme values at the positive and negative tolerance limits—yields the worst-case scenario (Table 1).

For the tightest tolerance, choose a smaller value for R₁ than for R_I and remember to account for the source characteristics of the input voltage, which must be able to drive the equivalent load of the network. When determining the final gain for the system, remember that the output impedance of the input voltage forms a voltage divider with the input impedance of the attenuation circuit. You can determine this load, the input resistance, as follows: For the differential configuration, R_IN=R₁ + R_I || (R₂/2). R_IN and C_IN form a highpass filter whose cutoff frequency, F_{-3 dB}, is 1/(2 x π x R_IN x C_IN). For the single-ended configuration, R_IN = R₁- R_I || (R₂ + R₁ || R_I). Input impedance for the attenuation network is:

where Z_IN is the input impedance and F is the input-signal frequency.

Deriving the highpass cutoff frequency for the single-ended case is less straightforward. To find the cutoff frequency, you must know the resistor values. Once you know them, you can solve the following equation for the cutoff frequency: Z_IN(f)=√ 2 x Z_IN(f=5000).

A spreadsheet is useful for determining the correct resistor values and system tolerances. Remember that resistors come in discrete values of resistance, and you must match rather than simply calculate them. A spreadsheet for determining custom gains and gain tolerances for single-ended and differential configurations of the MAX9705 is available at www.maxim-ic.com/ an-4559-supplement.